The volume of a sphere is increasing at a rate of $25\pi$ cubic meters per hour. At a certain instant, the volume is $\dfrac{32\pi}{3}$ cubic meters. What is the rate of change of the surface area of the sphere at that instant (in square meters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $25\pi$ (Choice B) B $\dfrac{32\pi}{9}$ (Choice C) C $\dfrac{25}{16}$ (Choice D) D $(\sqrt[3]{25\pi})^{^2}$ The surface area of a sphere with radius $r$ is $4\pi r^2$. The volume of a sphere with radius $r$ is $\dfrac{4}{3}\pi r^3$.
Solution: Setting up the math Let... $r(t)$ denote the sphere's radius at time $t$, $V(t)$ denote the sphere's volume at time $t$, and $S(t)$ denote the sphere's surface area at time $t$. We are given that $V'(t)=25\pi$, We are also given that $V(t_0)=\dfrac{32\pi}{3}$ for a specific time $t_0$. We want to find $S'(t_0)$. Relating the measures $S(t)$ and $r(t)$ relate to each other through the formula for the surface area of a sphere: $S(t)=4\pi[r(t)]^2$ We can differentiate both sides to find an expression for $S'(t)$ : $S'(t)=8\pi r(t)r'(t)$ $V(t)$ and $r(t)$ relate to each other through the formula for the volume of a sphere: $V(t)=\dfrac{4}{3}\pi[r(t)]^3$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=4\pi[r(t)]^2r'(t)$ Using the information to solve Let's plug ${V(t_0)}={\dfrac{32\pi}{3}}$ into the expression for $V(t_0)$ : $\begin{aligned} {V(t_0)}&=\dfrac43\pi[r(t_0)]^3 \\\\ {\dfrac{32\pi}{3}}&=\dfrac43\pi[r(t_0)]^3 \\\\ 8&=[r(t_0)]^3 \\\\ {2}&={r(t_0)} \end{aligned}$ Let's plug ${V'(t_0)}={25\pi}$ and ${r(t_0)}={2}$ into the expression for $V'(t_0)$ : $\begin{aligned} {V'(t_0)}&=4\pi[{r(t_0)}]^2r'(t_0) \\\\ {25\pi}&=4\pi({2})^2r'(t_0) \\\\ C{\dfrac{25}{16}}&=C{r'(t_0)} \end{aligned}$ Now let's plug ${r(t_0)}={2}$ and $C{r'(t_0)}=C{\dfrac{25}{16}}$ into the expression for $S'(t_0)$ : $\begin{aligned} S'(t_0)&=8\pi{r(t_0)}C{r'(t_0)} \\\\ &=8\pi({2})\left(C{\dfrac{25}{16}}\right) \\\\ &=25\pi \end{aligned}$ In conclusion, the rate of change of the surface area of the sphere at that instant is $25\pi$ square meters per hour. Since the rate of change is positive, we know that the surface area is increasing.